ta có A=\(\frac{2017^{2017}+1}{2017^{2018}+1}\)=> 2017A =\(\frac{2017^{2018}+2017}{2017^{2018}+1}=1+\frac{2016}{2017^{2018}+1}\)(1)
B=\(\frac{2017^{2018}+1}{2017^{2019}+1}\)=> 2017B =\(\frac{2017^{2019}+2017}{2017^{2019}+1}=1+\frac{2016}{2017^{2019}+1}\)(2)
So sánh (1)với (2) ta thấy 2017A>2017B
=>A>B
Vậy A>B
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{2017^{2018}+1}{2017^{2019}+1}< \frac{2017^{2018}+1+2016}{2017^{2019}+1+2016}=\frac{2017^{2018}+2017}{2017^{2017}+2017}=\frac{2017\left(2017^{2017}+1\right)}{2017\left(2017^{2016}+1\right)}=A\)
\(\Rightarrow\)\(B< A\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
Vi B = 2017^2019 > 2017^2018
=> B = 2017^2018 + 1/ 2017^2019 < 1 chon m = 2016
Ta co: 2017^2018 + 1 + 2016/ 2017^ 2019 + 1 + 2016
=> B < 2017^2018 + 2016/ 2017^2019 + 2016 = 2017 . 1 + 2017^ 2017 . 2017/ 2017 .1 + 2017^2018 . 1
=> B < 2017 . ( 2017^2017 + 1 )/ 2017 . ( 2017^ 2018 . 1 ) = 2017^2017 +1 / 2017^2018 +1 = A
=> B < A
Vay B < A